 In this example we've got a couple of different things going on. One is we want an equation of a line in standard form, which we've done before. What's different though is we want a line that's perpendicular to this given line. And then we've got our point, 6 and 9. So let's figure out some information so that we can get it in a point slope form, which we can then adjust into standard form. So we know the point that this line passes through is 6, 9. The slope of the given line is 3, since we have a line in slope intercept form. The perpendicular slope is the opposite reciprocal. So since 3 is positive, the perpendicular slope is going to be negative, and then the reciprocal of 3 is one-third. So now we can use this information, the perpendicular slope, and the point that we are given. And we can put that into point slope form, and from there we'll get standard form using some algebra work. First I'll take the point and that perpendicular slope, we'll put it into point slope form. Next we'll move towards slope intercept form, and the way we'll do that, first we'll distribute negative one-third times x, and negative one-third times negative 6. That leaves y minus 9 is negative one-third x. And now negative one-third times negative 6, that should be positive 2, add 9 to both sides. So now we have slope intercept form. Next our job is to turn that into standard form. So again we want the x's and the y's to live on the same side of the equal sign. So we'll add negative one, pardon, add positive one-third x to both sides. And then lastly to get standard form, we need each coefficient to be an integer. Since the coefficient out front of the x is one-third, I'll take and multiply both sides of this equation by 3. And then distribute that 3 in. 3 times one-third x is just x, 3 times y is 3y, and that equals 3 times 11, which is 33. And so there's your final answer in standard form.